Question: Simplify the following expression and state the condition under which the simplification is valid. $n = \dfrac{-7y^2 - 14y + 336}{-y^3 - 6y^2 + 16y}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ n = \dfrac {-7(y^2 + 2y - 48)} {-y(y^2 + 6y - 16)} $ $ n = \dfrac{7}{y} \cdot \dfrac{y^2 + 2y - 48}{y^2 + 6y - 16} $ Next factor the numerator and denominator. $ n = \dfrac{7}{y} \cdot \dfrac{(y + 8)(y - 6)}{(y + 8)(y - 2)}$ Assuming $y \neq -8$ , we can cancel the $y + 8$ $ n = \dfrac{7}{y} \cdot \dfrac{y - 6}{y - 2}$ Therefore: $ n = \dfrac{ 7(y - 6)}{ y(y - 2)}$, $y \neq -8$